In order to solve problems like this, you will need to understand prime factorization. Review this concept, if needed, before continuing. We will look at 3 common ways in which a polynomial can be factored: grouping, substitution, and using Identities.

Factor \( 25x^2 - 20x + 4 \).

Factor \( 16x^2 - 24xy + 9y^2 \).

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Factoring trinomials of the form \( x^2 + (a+b)x + ab = (x+a)(x+b): \)
This approach is also known as factorization by observation.

In cases where we don't have a perfect square of the form \( (x+y)^2 \) or \( (x-y)^2 \), but the leading coefficient of \( x \) is 1, we can try to find \(a \) and \( b\) such that \( a \cdot b\) is equal to the constant term, and \( a + b \) is equal to the coefficient of \( x \).

In both of these factorizations, the leading term will be \(2x \cdot x = 2x^2 \) and the constant term will be \( 1 \cdot 2 = 2 \), So we don't have to worry about them. Let's turn our attention to getting the correct middle term \( 4x \).

In the factorization \( (2x + 2)(x + 1) \), the middle term would be \( (2x \cdot 1) + (2 \cdot x) = 4x \), which is what we want. So the factorization that gives us the correct expansion is \( (2x + 2)(x + 1) \). \(_\square\)

There are many different combinations we can try for the factorization. But one thing to notice is that since the middle term is \( -11y \), we need to use factors that can get us to the number \( -11 \): \( 2y \times (-6) \) and \(y \times 1 \) would give us the middle term we want.

Since we want \( 2y \) to be multiplied by \( -6 \) to give us the middle term, the \( -6 \) must exist in a different set of parentheses from \( 2y \) so that it can be distributed to and multiplied by \( 2y \).

Factoring polynomials in this way involves some amount of guessing and checking. You can greatly improve your speed at this process by using your number sense to figure out which combinations of numbers will successfully get you the middle term that you want.

Factoring polynomials

Now that you're familiar with the different ways of factoring polynomials, let's work on some problems.

Given that \(6y^2=2014\), find the value of \[\frac{9(y^4+6y^3+9y^2)}{y^2+6y+9}. \]